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two particles of equal mass m go around a circle of radius R under the influence of their mutual gravitational attraction.the speed of each particle with respect to their center of mass will be??

bijitesh choudhury , 12 Years ago

Grade 12th Pass

7 Answers

rohit yadav

with respect to the center of mass the speed of each particle is (Gm/4R)^1/2 as their is no external force acting on the system so velocity of the center of mass is zero velocity of each particle is (Gm/4R)^1/2.

Last Activity: 12 Years ago

Chetan Mandayam Nayakar

mv²/R=Gm²/(2R)²

Last Activity: 12 Years ago

Yogita Bang

The gravitational force of attraction between the two masses is

F = Gm²/(2R)²

This force provides the necessary centripetal force

mv²/R = Gm²/4R²

v² = Gm/4R

v = √(Gm/4R)

Last Activity: 12 Years ago

gauhar singh

The Speed will be equal to their velocities in inertial frame itself as there centre of mass will be stationary.

Last Activity: 12 Years ago

Ojas Suvarnakar

Here center of mass will be stationary because both particles are having same mass,Therefore ,Speed of the particle will be as equal as critical velocity of the particle...Therefore answer is √GM/R

Last Activity: 8 Years ago

Aniket lohar

Two particles of equal mass(m) go around a circle of radius R under the action of their mutual gravitational attraction.Here center of mass is stationary because both particles are having same mass.1/2√(GM/R)

Last Activity: 8 Years ago

Yug Dedhia

Hello this is master Yug Dedhia

Here,using theory of gravitation

Therefore,mv^2/ r = Gm^2/(2R)^2

Thus the answer is v=√(Gm/4R)

Thanks

Last Activity: 6 Years ago

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